Re: Clifford Freaking Algebra

Thanks, great.

Yes, it seems there are two ways about this. As you point out, apparently the basis on page 209 is the standard physics way of presenting it, with

, even though that is not Penrose's own preference as he states in note 10 of chapter 24. I guess the two Clifford Algebra Introductions I found on the net (see above) show the mathematical way of presenting the thing with

. Perhaps Penrose presents the standard physics way because it flows more directly from the discussion of quaternions, and because two reflections equal a

rotation, which equates to -1 for a spinor object. But, the mathematical way seems a bit nicer to me. as I explain below, basically repeating something from the first web site in my initial post.

If you have a two dimensional space with two orthonormal basis vectors

and

, with

and

, two general elements

and

of the Clifford algebra could be represented as follows

where

are scalars

The product

is found by multiplying the components distributively and simplifying based on

so, for example

So,

where the scalars

are

However, if you start out saying

, then

Which is not quite as symmetrical, and lead me to think that maybe the two ways were not equivalent